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I Area of sector by integration

  1. Mar 5, 2016 #1
    I am bothered on how the area of circular sector was derived using integration. I have a solution but I don't know how it will go and come up with the formula:

    A = 1/2 sr and A = 1/2 r2 θ

    Can someone show me the solution for the derivation of the two formula?
     
  2. jcsd
  3. Mar 5, 2016 #2

    blue_leaf77

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    A basic question, what do you know about an integral?
     
  4. Mar 5, 2016 #3

    SteamKing

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    It's an elementary proof if you use polar coordinates.

    Examine the small sector in the figure below:

    gaussianGraph.gif

    You can assume for a small angle dθ that the sector is roughly an isosceles triangle, where each side measures r units and the base is r ⋅ dθ units long.

    The area of this triangle is dA ≈ (1/2) ⋅ r × r ⋅ dθ = r2 dθ / 2, using the formula for the area of a triangle.

    If you want to find the area A of a circular sector between the angles θ1 and θ2, then

    $$A = \int_{θ1}^{θ2} \,dA = \int_{θ1}^{θ2} \frac{1}{2} r^2\,dθ = \frac{1}{2} r^2 ⋅ (θ2 - θ1)$$

    r ⋅ (θ2 - θ1) is also equivalent to the arc length s, and ##A = \frac{1}{2} r ⋅ s##

    Q.E.D.
     
  5. Mar 5, 2016 #4
    I have done it by means of basic integration of plane areas. I divide the figure into two parts as A1 and A2 for computing the area. 272026.image3.jpg

    It is somehow like this (I can't find a picture of my solution). I draw a triangle from A perpendicular to CB. Now, the sector is divided into a triangle and segment. The triangle represents A1 and segment represents A2. Can you show me your solution in my way of derivation?
     
  6. Mar 5, 2016 #5

    SteamKing

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    No, because that's a really difficult way to do this problem.

    There's nothing wrong with switching to integration using polar coordinates when that method makes the calculation much easier.
     
  7. Mar 5, 2016 #6
    I am finish in integrating it with values.. I am stuck on how the integration will cancel out to come up with only 1/2 r2 θ for the area.

    What do you mean by polar coordinates? I think I work with cartesian coordinates in integrating?
     
  8. Mar 5, 2016 #7

    SteamKing

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    r is a constant with respect to theta. What happens when you integrate constants?
    I mean polar coordinates:

    https://en.wikipedia.org/wiki/Polar_coordinate_system

    If you haven't studied polar coordinates, there is a certain mathematical background which is lacking in your education. You should correct this if you are going to study calculus.
     
  9. Mar 5, 2016 #8
    I know how to integrate constants. I just don't know how to simplify my solution to 1/2 r2 θ.

    We had study about polar coordinates and I know it, I just don't know how to make equations for the integration process.
     
  10. Mar 5, 2016 #9

    SteamKing

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    You'll have to show your work where you got stuck.
    I show you how and explained it in Post #3.
     
  11. Mar 5, 2016 #10
    I got this part: πr2 / 4 - r2/2 arcsin (cos θ)

    Can you show me how, please?
     
  12. Mar 5, 2016 #11
    This is my solution:
     

    Attached Files:

  13. Mar 5, 2016 #12

    SteamKing

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    I'm not even sure what this expression is supposed to represent.

    If it's supposed to be the area of a circular sector subtending an angle of 45°, it's way more complicated than it needs to be (and I haven't checked for any mistakes).

    https://en.wikipedia.org/wiki/Circular_sector
     
  14. Mar 5, 2016 #13
    Have you checked my solution in the attached file?

    For the formula A = 1/2 sr, how it was derived using integration? Sorry for so many questions, I just want to clear up my mind.
     
  15. Mar 5, 2016 #14

    SteamKing

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    I looked at it. It's way too complicated to spend time checking, especially when I have shown you in detail how to obtain a much simpler formula for the area of a circular sector, and showed you documentary proof of the validity of the simple formula.
    Again, please refer to Post #3. I showed how this formula is derived from the alternate formula for the area of a circular sector.

    If you are not going to read the posts I put up, there's no point in continuing this thread.
     
  16. Mar 5, 2016 #15
    How come my picture of my integration of sector can be drawn like this? Is this also the same with mine?
     
  17. Mar 6, 2016 #16

    SteamKing

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    All circular sectors look alike.

    The dA in the figure is actually drawn for the case where r is not a constant but instead varies with the angle θ.

    For a circular sector, r is constant, and integrating r and θ is not necessary to find the area of the sector.
     
  18. Mar 6, 2016 #17
    How about the curve in the part of the sector? If I'm not mistaken, the figure you posted is somehow a triangle rather than a sector with curve part.
    Where is it?
     
  19. Mar 6, 2016 #18

    SteamKing

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    In the limit, as dθ → 0, then the arc portion of the sector is naturally going to appear more and more like a straight line segment, which is why the area of this tiny sector can be approximated using the formula for the area of a triangle.

    To find the area of circular sector ABC, as shown in Post #4, that sector would be subdivided into many little slivers each sweeping out an angle dθ. The integration process adds up the areas dA of these segments as dθ → 0, producing a finite value A for the area of sector ABC.
     
  20. Mar 7, 2016 #19
    In general, the area of a sector bounded by a curve ##C## defined by polar radius ##\rho(\theta)##, such that ##C = \rho([\theta_1,\theta_2])## , is ## A= \frac{1}{2} \int_{\theta_1}^{\theta_2} \rho^2(\theta) \ d\theta ## . In your case, ##\rho(\theta)## is constant and equal to ##r## since your sector is circular.
     
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